How do you solve using the completing the square method #3x^2-4x-2=0#?

1 Answer
Shwetank Mauria
Aug 6, 2016

#x=(-2-sqrt10)/3# or #x=(-2+sqrt10)/3#

Explanation:

Let us divide each term of #3x^2-4x-2=0# by #3#, we get

#x^2-4/3x-2/3=0#

Now recalling the identity #(x-a)^2=x^2-2ax+a^2# and comparing it with #x^2-4/3x#, we need to add and subtract #(4/(3xx2))^2# to complete square. Hence #x^2-4/3x-2/3=0# is

#hArrx^2-2xx4/6x+(4/6)^2-(4/6)^2-2/3=0# or

#(x+4/6)^2-(2/3)^2-2/3=0# or

#(x+2/3)^2-4/9-6/9=0# or

#(x+2/3)^2-10/9=0# or

#(x+2/3)^2-(sqrt10/3)^2=0# or

#(x+2/3+sqrt10/3)(x+2/3-sqrt10/3)=0#

Hence, #x+2/3+sqrt10/3=0# or #x+2/3-sqrt10/3=0#

i.e. #x=(-2-sqrt10)/3# or #x=(-2+sqrt10)/3#

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